ma 201: method of separation of variables finite vibrating … · · 2016-10-26ma 201: method of...

MA 201: Method of Separation of Variables

Finite Vibrating String ProblemLecture - 11

MA201(2016): PDE

IBVP for Vibrating string with no external forces

• We consider the problem in a computational domain

(x , t) ∈ [0, L]× [0,∞)

• The IBVP under consideration consists of the following:

• The governing equation:

utt = c2uxx , (x , t) ∈ (0, L)× (0,∞). (1)

The boundary conditions for all t > 0:

u(0, t) = 0, u(L, t) = 0. (2)

The initial conditions for 0 ≤ x ≤ L are

u(x , 0) = φ(x), ut(x , 0) = ψ(x). (3)

MA201(2016): PDE

Separation of variables method

• Method of separation of variables for PDE:a convenient and powerful solution technique of partial differentialequations.

• The main idea of this method is toconvert the given partial differential equation into several ordinarydifferential equations and then obtain the solutions by familiar solutiontechniques.

• The success of this method stems from the fact thata function can be expanded in a series in terms of certainelementary/special functions.

• The basic idea of this method is thatthe solution is assumed to consist of the product of two or morefunctions:

• Each function being the function of one independent variableonly.

MA201(2016): PDE

Separation of variables method (Contd.)

• The number of functions involved dependson the number of independent variables.

• As for exampleTry solving an equation for u = u(x , y).

• We will assume a solution in the form u(x , y ) = X (x)Y (y )where X is a function of x only and Y is a function of y only.

• Substituting this solution in the given equationwe will have a pair of ODEs.

• Note thatThis method can be used only for bounded domains so that boundaryconditions are prescribed.

MA201(2016): PDE

Boundary and initial conditions

• If the boundary conditions are prescribed in terms of somevalues of the function on the boundaries,then these conditions are called Dirichlet conditions.

• and the corresponding BVPs are called Dirichlet boundaryvalue problems.

• If the boundary conditions are prescribed in terms of somevalues of the derivatives on the boundariesthen these conditions are called Neumann conditions.

• and the corresponding BVPs are called Neumann boundaryvalue problems.

• If the boundary conditions for a specific problem contain bothtypes,then these conditions are called mixed or Robin conditions and thecorresponding BVPs are called Robin boundary value problem.

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)

• Recall the wave equation

utt − c2uxx = 0. (4)

• Assume a solution of the form

u(x , t) = X (x)T (t). (5)

• Here, X (x) is function of x alone and T (t) is a function of talone.

• Substituting (5) in equation (4)

XT ′′ = c2X ′′T . (6)

MA201(2016): PDE


• Separating the variables

X ′′

X=

T ′′

c2T.

• Here the left side is a function of x and the right side is afunction of t.

• The equality will hold only if both are equal to a constant, say,k.

• We get two differential equations as follows:

X ′′ − kX = 0, (7a)

T ′′ − c2kT = 0. (7b)

MA201(2016): PDE


• Since k is any constant,it can be zero, orit can be positive, orit can be negative.

• Consider all the possibilities andexamine what value(s) of k lead to a non-trivial solution.

MA201(2016): PDE


Case I: k = 0.

• In this case the equations (7) reduce to

X ′′ = 0, and T ′′ = 0

• Giving rise to solutions

X (x) = Ax + B, T (t) = Ct + D.

• But the solution u(x , t) = X (x)T (t) is a trivial solutionif it has to satisfy the boundary conditions u(0, t) = u(L, t) = 0,

• This case of k = 0 is rejected sinceit gives rise to trivial solutions only.

MA201(2016): PDE


Case II: k > 0, let k = λ2 for some λ > 0.

• In this case the equations (7) reduce to the equations

X ′′ − λ2X = 0, and T ′′ − c2λ2T = 0


X (x) = Aeλx + Be−λx ,

T (t) = Cecλt + De−cλt .

• Therefore

u(x , t) = (Aeλx + Be−λx)(Cecλt + De−cλt).

MA201(2016): PDE


• Using boundary condition u(0, t) = 0,

A+ B = 0, B = −A.

• Using boundary condition u(L, t) = 0,

(AeλL + Be−λL)(Cecλt + De−cλt) = 0.

• The t part of the solution cannot be zero as it will lead to a trivialsolution.

• Then we must have

A(eλL − e−λL) = 0,

• Which leads to A = 0 as λ 6= 0.

• k > 0 also gives rise to trivial solution:k > 0 is also rejected.

MA201(2016): PDE


Case III: k < 0, let k = −λ2 for some λ > 0.

• In this case equations (7) reduce to the equations

X ′′ + λ2X = 0 and T ′′ + c2λ2T = 0


X (x) = A cosλx + B sinλx ,

T (t) = C cos(cλt) + D cos(cλt).

• Hence

u(x , t) = (A cosλx + B sinλx)(C cos(cλt) + D sin(cλt)).

MA201(2016): PDE


• Using boundary condition u(0, t) = 0, A = 0.

• Using boundary condition u(L, t) = 0,

B sinλL = 0.

• B 6= 0 as that will lead to a trivial solution.

• Hence we must have

sinλL = 0

• which gives us

λ = λn =nπ

L, n = 1, 2, 3, . . .

MA201(2016): PDE


• These λn’s are called eigenvalues and note thatcorresponding to each n there will be an eigenvalue.

• Accordingly, the solution is

u(x , t) = (A cosλx + B sinλx)(C cos(cλt) + D sin(cλt))

= sinλnx(BC cos(cλnt) + BD sin(cλnt))

= sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

= un(x , t).

• The solution corresponding to each eigenvalue is called aneigenfunction

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• Since the wave equation is linear and homogeneous, any linearcombination will also be a solution

• Hence, we can expect the solution in the following form:

u(x , t) =

∞∑

n=1

un(x , t)

=∞∑

n=1

sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

, (8)

• providedi. An and Bn are determined uniquely andii. each of the resulting series for those coefficients converges, andiii. the limit of the series is twice continuously differentiable with respectto x and t so that it satisfies the equation utt − c2uxx = 0,

MA201(2016): PDE


• What is the idea?

• First, we assume that (!)(i) the infinite series converges for some An and Bn

(ii) term-wise differentiation with respect to t is possible andit converges

• Next, we calculate An and Bn using the given initial conditions.

• Once, both the coefficients An and Bn are calculated, we thenprove that the series actually holds following properties

i. the resulting series for those coefficients converge, and

ii. the limit of the series is twice continuously differentiable with respectto x and t, and it satisfies the partial differential equation.

MA201(2016): PDE


• Using the initial condition u(x , 0) = φ(x)

φ(x) =

∞∑

n=1

An sinnπx

L. (9)

• This series can be recognized as the half-range sine expansion of afunction φ(x) defined in the range (0, L). What is the assumption on φfor the convergence in (0, L)?

• An can be obtained by multiplying equation (9) by sin nπx

Land

integrating with respect to x from 0 to L.

• Therefore

An =2

L

∫ L

0

φ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (10)

• Here, we have used the fact that

∫

L

−L

sinnπx

Lsin

nπx

Ldx = L.

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• To use the other initial condition ut(x , 0) = ψ(x), we need todifferentiate (8) w.r.t. t to get

ut(x , t) =∞∑

n=1

sinnπx

L

(nπc

L

) [

−An sinnπct

L+ Bn cos

nπct

L

]

.

• Then

ψ(x) =∞∑

n=1

Bn

nπc

Lsin

nπx

L.

• Similarly

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (11)

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• Now, consider the infinite series

∞∑

n=1

un(x , t) =∞∑

n=1

sinnπx

L

[

An cosnπct

L+ Bn sin

nπct

L

]

, (12)

with

•

An =2

L

∫ L

0

φ(x) sinnπx

Ldx , n = 1, 2, 3, . . . (13)

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

Ldx , n = 1, 2, 3, . . . . (14)

• Clearly, the series satisfies both initial and boundary conditions.

• In fact, we claim that it is the solution of the finite vibrating stringproblem. How?

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• Consider following infinite series

w1(x , t) =∞∑

n=1

An cosnπct

Lsin

nπx

L, (15)

w2(x , t) =

∞∑

n=1

Bn sinnπct

Lsin

nπx

L. (16)

• Using the trigonometric identity

sinnπx

Lcos

nπct

L=

1

2sin

nπ

L(x − ct) +

1

2sin

nπ

L(x + ct) (17)

• We write

w1(x , t) =1

2

∞∑

n=1

An sinnπ

L(x − ct) +

1

2

∞∑

n=1

An sinnπ

L(x + ct)

=1

2

∞∑

n=1

An sinnπ

L(ξ) +

1

2

∞∑

n=1

An sinnπ

L(η),

ξ = x − ct ∈ R, η = x + ct ∈ R.

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• We know that the Fourier series

∞∑

n=1

An sinnπξ

Lwith An =

2

L

∫

L

0

φ(x) sinnπx

Ldx

converges to φ(ξ) in ξ ∈ (0, L). But, ξ ∈ R.

• Define the odd periodic extension of φ by

φo(x) = φ(x) if 0 < x < L,

φo(x) = −φ(−x) if − L < x < 0,

φo(x) = φ(x + 2L), for rest x ∈ R.

• From convergence of Fourier series, we conclude that the series

∞∑

n=1

An sinnπξ

Lwith An =

1

L

∫

L

−L

φo(x) sinnπx

Ldx =

2

L

∫

L

0

φo(x) sinnπx

Ldx = An

converges to φo(ξ) in [−L, L]. What about the convergence in (−∞,∞)?

MA201(2016): PDE

IBVP for Vibrating string with no external forces (Contd.)• Therefore, the infinite series

∞∑

n=1

An sinnπξ

Lwith An =

2

L

∫

L

0

φ(x) sinnπx

Ldx

converges to φo(ξ) in ξ ∈ (−∞,∞).

• As a consequence, we have

w1(x , t) =1

2φo(ξ) +

1

2φo(η)

• Assume that φ ∈ C 2[0, L]. What about the smoothness of φo in(−∞,∞)?

• Then we obtain

∂2w1

∂t2− c2

∂2w1

∂x2= 0.

MA201(2016): PDE


• Similarly, under the assumption ψ ∈ C 2[0, L], we have

∂2w2

∂t2− c2

∂2w2

∂x2= 0.

• Hence,

∂2

∂t2(w1 + w2)− c2

∂2

∂x2(w1 + w2) = 0.

• Recall

w1(x , t) =∞∑

n=1

An cosnπct

Lsin

nπx

L,

w2(x , t) =

∞∑

n=1

Bn sinnπct

Lsin

nπx

L.

MA201(2016): PDE

Formal Solution of the Finite Vibrating String Problem

• The solution is given by

u(x , t) = w1(x , t) + w2(x , t)

=∞∑

n=1

An cosnπct

Lsin

nπx

L+

∞∑

n=1

Bn sinnπct

Lsin

nπx

L

• with

An =2

L

∫ L

0

φ(x) sinnπx

Ldx , n = 1, 2, 3, . . .

Bn =2

nπc

∫ L

0

ψ(x) sinnπx

Ldx , n = 1, 2, 3, . . . .

• Is the solution unique?

MA201(2016): PDE

ma 201: method of separation of variables finite vibrating … · · 2016-10-26ma 201: method of...

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